Integrand size = 25, antiderivative size = 172 \[ \int \frac {\cot ^6(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=-\frac {\arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{\sqrt {a} f}-\frac {\left (15 a^2+40 a b+33 b^2\right ) \cot (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{15 (a+b)^3 f}+\frac {(5 a+9 b) \cot ^3(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{15 (a+b)^2 f}-\frac {\cot ^5(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{5 (a+b) f} \]
-arctan(a^(1/2)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)^(1/2))/f/a^(1/2)-1/15*(15* a^2+40*a*b+33*b^2)*cot(f*x+e)*(a+b+b*tan(f*x+e)^2)^(1/2)/(a+b)^3/f+1/15*(5 *a+9*b)*cot(f*x+e)^3*(a+b+b*tan(f*x+e)^2)^(1/2)/(a+b)^2/f-1/5*cot(f*x+e)^5 *(a+b+b*tan(f*x+e)^2)^(1/2)/(a+b)/f
Time = 3.34 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.16 \[ \int \frac {\cot ^6(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=-\frac {\arctan \left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b-a \sin ^2(e+f x)}}\right ) \sqrt {a+2 b+a \cos (2 e+2 f x)} \sec (e+f x)}{\sqrt {2} \sqrt {a} f \sqrt {a+b \sec ^2(e+f x)}}-\frac {(a+2 b+a \cos (2 (e+f x))) \csc (e+f x) \left (23 a^2+60 a b+45 b^2-\left (11 a^2+26 a b+15 b^2\right ) \csc ^2(e+f x)+3 (a+b)^2 \csc ^4(e+f x)\right ) \sec (e+f x)}{30 (a+b)^3 f \sqrt {a+b \sec ^2(e+f x)}} \]
-((ArcTan[(Sqrt[a]*Sin[e + f*x])/Sqrt[a + b - a*Sin[e + f*x]^2]]*Sqrt[a + 2*b + a*Cos[2*e + 2*f*x]]*Sec[e + f*x])/(Sqrt[2]*Sqrt[a]*f*Sqrt[a + b*Sec[ e + f*x]^2])) - ((a + 2*b + a*Cos[2*(e + f*x)])*Csc[e + f*x]*(23*a^2 + 60* a*b + 45*b^2 - (11*a^2 + 26*a*b + 15*b^2)*Csc[e + f*x]^2 + 3*(a + b)^2*Csc [e + f*x]^4)*Sec[e + f*x])/(30*(a + b)^3*f*Sqrt[a + b*Sec[e + f*x]^2])
Time = 0.45 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.09, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 4629, 2075, 382, 25, 445, 445, 27, 291, 216}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {\cot ^6(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\tan (e+f x)^6 \sqrt {a+b \sec (e+f x)^2}}dx\) |
| \(\Big \downarrow \) 4629 |
| \(\displaystyle \frac {\int \frac {\cot ^6(e+f x)}{\left (\tan ^2(e+f x)+1\right ) \sqrt {a+b \left (\tan ^2(e+f x)+1\right )}}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 2075 |
| \(\displaystyle \frac {\int \frac {\cot ^6(e+f x)}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 382 |
| \(\displaystyle \frac {\frac {\int -\frac {\cot ^4(e+f x) \left (4 b \tan ^2(e+f x)+5 a+9 b\right )}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{5 (a+b)}-\frac {\cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{5 (a+b)}}{f}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {-\frac {\int \frac {\cot ^4(e+f x) \left (4 b \tan ^2(e+f x)+5 a+9 b\right )}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{5 (a+b)}-\frac {\cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{5 (a+b)}}{f}\) |
| \(\Big \downarrow \) 445 |
| \(\displaystyle \frac {-\frac {-\frac {\int \frac {\cot ^2(e+f x) \left (15 a^2+40 b a+33 b^2+2 b (5 a+9 b) \tan ^2(e+f x)\right )}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{3 (a+b)}-\frac {(5 a+9 b) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{3 (a+b)}}{5 (a+b)}-\frac {\cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{5 (a+b)}}{f}\) |
| \(\Big \downarrow \) 445 |
| \(\displaystyle \frac {-\frac {-\frac {-\frac {\int \frac {15 (a+b)^3}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{a+b}-\frac {\left (15 a^2+40 a b+33 b^2\right ) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{a+b}}{3 (a+b)}-\frac {(5 a+9 b) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{3 (a+b)}}{5 (a+b)}-\frac {\cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{5 (a+b)}}{f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {-\frac {-\frac {-15 (a+b)^2 \int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)-\frac {\left (15 a^2+40 a b+33 b^2\right ) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{a+b}}{3 (a+b)}-\frac {(5 a+9 b) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{3 (a+b)}}{5 (a+b)}-\frac {\cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{5 (a+b)}}{f}\) |
| \(\Big \downarrow \) 291 |
| \(\displaystyle \frac {-\frac {-\frac {-15 (a+b)^2 \int \frac {1}{\frac {a \tan ^2(e+f x)}{b \tan ^2(e+f x)+a+b}+1}d\frac {\tan (e+f x)}{\sqrt {b \tan ^2(e+f x)+a+b}}-\frac {\left (15 a^2+40 a b+33 b^2\right ) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{a+b}}{3 (a+b)}-\frac {(5 a+9 b) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{3 (a+b)}}{5 (a+b)}-\frac {\cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{5 (a+b)}}{f}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {-\frac {-\frac {-\frac {\left (15 a^2+40 a b+33 b^2\right ) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{a+b}-\frac {15 (a+b)^2 \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{\sqrt {a}}}{3 (a+b)}-\frac {(5 a+9 b) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{3 (a+b)}}{5 (a+b)}-\frac {\cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{5 (a+b)}}{f}\) |
(-1/5*(Cot[e + f*x]^5*Sqrt[a + b + b*Tan[e + f*x]^2])/(a + b) - (-1/3*((5* a + 9*b)*Cot[e + f*x]^3*Sqrt[a + b + b*Tan[e + f*x]^2])/(a + b) - ((-15*(a + b)^2*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]])/Sqr t[a] - ((15*a^2 + 40*a*b + 33*b^2)*Cot[e + f*x]*Sqrt[a + b + b*Tan[e + f*x ]^2])/(a + b))/(3*(a + b)))/(5*(a + b)))/f
3.5.14.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Subst [Int[1/(c - (b*c - a*d)*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]
Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_) , x_Symbol] :> Simp[(e*x)^(m + 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/ (a*c*e*(m + 1))), x] - Simp[1/(a*c*e^2*(m + 1)) Int[(e*x)^(m + 2)*(a + b* x^2)^p*(c + d*x^2)^q*Simp[(b*c + a*d)*(m + 3) + 2*(b*c*p + a*d*q) + b*d*(m + 2*p + 2*q + 5)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[ b*c - a*d, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, d, e, m, 2, p, q, x]
Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_ .)*((e_) + (f_.)*(x_)^2), x_Symbol] :> Simp[e*(g*x)^(m + 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(a*c*g*(m + 1))), x] + Simp[1/(a*c*g^2*(m + 1)) Int[(g*x)^(m + 2)*(a + b*x^2)^p*(c + d*x^2)^q*Simp[a*f*c*(m + 1) - e*(b*c + a*d)*(m + 2 + 1) - e*2*(b*c*p + a*d*q) - b*e*d*(m + 2*(p + q + 2) + 1)*x^ 2, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p, q}, x] && LtQ[m, -1]
Int[(u_)^(p_.)*(v_)^(q_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Int[(e*x)^m*Expa ndToSum[u, x]^p*ExpandToSum[v, x]^q, x] /; FreeQ[{e, m, p, q}, x] && Binomi alQ[{u, v}, x] && EqQ[BinomialDegree[u, x] - BinomialDegree[v, x], 0] && ! BinomialMatchQ[{u, v}, x]
Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*((d_.)*tan[(e_.) + (f _.)*(x_)])^(m_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Sim p[ff/f Subst[Int[(d*ff*x)^m*((a + b*(1 + ff^2*x^2)^(n/2))^p/(1 + ff^2*x^2 )), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, d, e, f, m, p}, x] && Inte gerQ[n/2] && (IntegerQ[m/2] || EqQ[n, 2])
Leaf count of result is larger than twice the leaf count of optimal. \(1278\) vs. \(2(154)=308\).
Time = 6.78 (sec) , antiderivative size = 1279, normalized size of antiderivative = 7.44
-1/15/f/(a+b)^3/(-a)^(1/2)*(15*sin(f*x+e)^5*ln(4*(-a)^(1/2)*((b+a*cos(f*x+ e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/ (1+cos(f*x+e))^2)^(1/2)-4*sin(f*x+e)*a)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e)) ^2)^(1/2)*a^3*cos(f*x+e)+45*sin(f*x+e)^5*ln(4*(-a)^(1/2)*((b+a*cos(f*x+e)^ 2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+ cos(f*x+e))^2)^(1/2)-4*sin(f*x+e)*a)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2) ^(1/2)*a^2*b*cos(f*x+e)+45*sin(f*x+e)^5*ln(4*(-a)^(1/2)*((b+a*cos(f*x+e)^2 )/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+c os(f*x+e))^2)^(1/2)-4*sin(f*x+e)*a)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^ (1/2)*a*b^2*cos(f*x+e)+15*sin(f*x+e)^5*ln(4*(-a)^(1/2)*((b+a*cos(f*x+e)^2) /(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+co s(f*x+e))^2)^(1/2)-4*sin(f*x+e)*a)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^( 1/2)*b^3*cos(f*x+e)+23*cos(f*x+e)^6*(-a)^(1/2)*a^3+60*cos(f*x+e)^6*(-a)^(1 /2)*a^2*b+45*cos(f*x+e)^6*(-a)^(1/2)*a*b^2+15*sin(f*x+e)^5*ln(4*(-a)^(1/2) *((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+4*(-a)^(1/2)*((b+a *cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-4*sin(f*x+e)*a)*((b+a*cos(f*x+e)^2) /(1+cos(f*x+e))^2)^(1/2)*a^3+45*sin(f*x+e)^5*ln(4*(-a)^(1/2)*((b+a*cos(f*x +e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+4*(-a)^(1/2)*((b+a*cos(f*x+e)^2) /(1+cos(f*x+e))^2)^(1/2)-4*sin(f*x+e)*a)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e) )^2)^(1/2)*a^2*b+45*sin(f*x+e)^5*ln(4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1...
Leaf count of result is larger than twice the leaf count of optimal. 433 vs. \(2 (154) = 308\).
Time = 1.91 (sec) , antiderivative size = 987, normalized size of antiderivative = 5.74 \[ \int \frac {\cot ^6(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\text {Too large to display} \]
[-1/120*(15*((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cos(f*x + e)^4 + a^3 + 3*a^2* b + 3*a*b^2 + b^3 - 2*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cos(f*x + e)^2)*sqrt (-a)*log(128*a^4*cos(f*x + e)^8 - 256*(a^4 - a^3*b)*cos(f*x + e)^6 + 32*(5 *a^4 - 14*a^3*b + 5*a^2*b^2)*cos(f*x + e)^4 + a^4 - 28*a^3*b + 70*a^2*b^2 - 28*a*b^3 + b^4 - 32*(a^4 - 7*a^3*b + 7*a^2*b^2 - a*b^3)*cos(f*x + e)^2 - 8*(16*a^3*cos(f*x + e)^7 - 24*(a^3 - a^2*b)*cos(f*x + e)^5 + 2*(5*a^3 - 1 4*a^2*b + 5*a*b^2)*cos(f*x + e)^3 - (a^3 - 7*a^2*b + 7*a*b^2 - b^3)*cos(f* x + e))*sqrt(-a)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e)) *sin(f*x + e) + 8*((23*a^3 + 60*a^2*b + 45*a*b^2)*cos(f*x + e)^5 - (35*a^3 + 94*a^2*b + 75*a*b^2)*cos(f*x + e)^3 + (15*a^3 + 40*a^2*b + 33*a*b^2)*co s(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(((a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*f*cos(f*x + e)^4 - 2*(a^4 + 3*a^3*b + 3*a^2*b^2 + a*b ^3)*f*cos(f*x + e)^2 + (a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*f)*sin(f*x + e) ), 1/60*(15*((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cos(f*x + e)^4 + a^3 + 3*a^2* b + 3*a*b^2 + b^3 - 2*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cos(f*x + e)^2)*sqrt (a)*arctan(1/4*(8*a^2*cos(f*x + e)^5 - 8*(a^2 - a*b)*cos(f*x + e)^3 + (a^2 - 6*a*b + b^2)*cos(f*x + e))*sqrt(a)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/((2*a^3*cos(f*x + e)^4 - a^2*b + a*b^2 - (a^3 - 3*a^2*b)*cos(f*x + e)^2)*sin(f*x + e)))*sin(f*x + e) - 4*((23*a^3 + 60*a^2*b + 45*a*b^2)*cos (f*x + e)^5 - (35*a^3 + 94*a^2*b + 75*a*b^2)*cos(f*x + e)^3 + (15*a^3 +...
\[ \int \frac {\cot ^6(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\int \frac {\cot ^{6}{\left (e + f x \right )}}{\sqrt {a + b \sec ^{2}{\left (e + f x \right )}}}\, dx \]
\[ \int \frac {\cot ^6(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\int { \frac {\cot \left (f x + e\right )^{6}}{\sqrt {b \sec \left (f x + e\right )^{2} + a}} \,d x } \]
\[ \int \frac {\cot ^6(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\int { \frac {\cot \left (f x + e\right )^{6}}{\sqrt {b \sec \left (f x + e\right )^{2} + a}} \,d x } \]
Timed out. \[ \int \frac {\cot ^6(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\int \frac {{\mathrm {cot}\left (e+f\,x\right )}^6}{\sqrt {a+\frac {b}{{\cos \left (e+f\,x\right )}^2}}} \,d x \]